Problem 44 - Chemistry: The Central Science

Question

The first-order rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}, 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\), at \(70^{\circ} \mathrm{C}\) is \(6.82 \times\) \(10^{-3} \mathrm{~s}^{-1}\). Suppose we start with \(0.0250 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g})\) in a volume of \(2.0 \mathrm{~L}\). (a) How many moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) will remain after \(5.0 \mathrm{~min}\) ? (b) How many minutes will it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to \(0.010\) mol? (c) What is the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(70{ }^{\circ} \mathrm{C}\) ?

Step-by-Step Solution

Answer

(a) There will be \(0.0110 \, \text{mol}\) of N2O5 remaining after 5.0 minutes. (b) It will take \(4.44 \, \text{min}\) for the quantity of N2O5 to drop to 0.010 mol. (c) The half-life of N2O5 at \(70^{\circ} \mathrm{C}\) is \(1.69 \, \text{min}\).

Step 1: Convert minutes to seconds

Because the rate constant is given in s^-1, we must convert the given time of 5.0 minutes into seconds: \( t = 5.0 \times 60 = 300 \mathrm{~s} \)

Step 2: Insert known values and calculate

Using the given values and the first-order rate equation, we can calculate the remaining moles of N2O5 after 5.0 minutes: \( N_t = (0.0250) e^{-(6.82 \times 10^{-3})(300)} \) \( N_t = 0.0250 e^{-2.046} \) \( N_t = 0.0110 \, \text{mol} \) So, there will be 0.0110 mol of N2O5 remaining after 5.0 minutes. b) Time for N2O5 to drop to 0.010 mol To find the time for the amount of N2O5 to drop to 0.010 mol, we can use the first-order rate equation again: \( t = \dfrac{\ln{(\frac{N_t}{N_0})}}{-k} \)

Step 1: Insert known values and calculate

Using the given values and the first-order rate equation, we can calculate the time it takes for N2O5 to drop to 0.010 mol: \( t = \dfrac{\ln{(\frac{0.010}{0.0250})}}{-6.82 \times 10^{-3}} \) \( t = 266.22 \, \text{s} \)

Step 2: Convert seconds to minutes

Convert the calculated time in seconds to minutes: \( t = \dfrac{266.22}{60} = 4.44 \, \text{min} \) So it will take 4.44 minutes for the quantity of N2O5 to drop to 0.010 mol. c) Half-life of N2O5 at 70°C The half-life of a first-order reaction can be found with the following equation: \( t_{1/2} = \dfrac{\ln{2}}{k} \)

Step 1: Insert known values and calculate

Using the given rate constant and the equation for the half-life, we can calculate the half-life of N2O5 at 70°C: \( t_{1/2} = \dfrac{\ln{2}}{6.82 \times 10^{-3}} \) \( t_{1/2} = 101.63 \, \text{s} \)

Step 2: Convert seconds to minutes

Convert the calculated half-life in seconds to minutes: \( t_{1/2} = \dfrac{101.63}{60} = 1.69 \, \text{min} \) So, the half-life of N2O5 at 70°C is 1.69 minutes.

Key Concepts

Chemical Kinetics

Chemical kinetics is the branch of chemistry that deals with the rate of chemical reactions and the mechanisms by which they occur. Think of the rate as how fast a car is moving, but instead of a car, we have molecules reacting to form products. For any reaction, the rate can be expressed as the change in concentration of a reactant or product over time.

The rate of a chemical reaction is influenced by several factors, including the concentration of reactants, temperature, presence of a catalyst, and the specific properties of the reactants and products. In our exercise, the rate constant for the decomposition of \( \mathrm{N}_{2}\mathrm{O}_{5} \) illustrates the specific rate at which this reactant is transforming into its products at a given temperature of \(70^\circ \mathrm{C}\). It's crucial for students to understand that this 'rate constant' is unique to each reaction and its conditions, acting as a fingerprint for that reaction's speed under those circumstances.

First-Order Reaction

A first-order reaction is one where the rate is directly proportional to the concentration of one reactant. It's like a self-service buffet; the amount of food you grab (the rate you eat) depends only on your hunger (the concentration of the reactant). Mathematically, for a reaction \( A \rightarrow Products \) with \( A \) being our reactant, the rate law is written as \( \text{rate} = k[A] \) where \( k \) is the first-order rate constant.

In our textbook example, the decomposition of \( \mathrm{N}_{2}\mathrm{O}_{5} \) follows a first-order rate equation. To determine how much \( \mathrm{N}_{2}\mathrm{O}_{5} \) remains after 5 minutes, we used the first-order rate equation incorporating the exponential function. This equation shows that the rate at which \( \mathrm{N}_{2}\mathrm{O}_{5}\) decomposes depends solely on its own initial concentration and not on the concentration of any other substances.

Half-Life

The half-life of a reaction, represented by \( t_{1/2} \), is the time it takes for half of the reactant to be consumed or for the concentration of the reactant to decrease to half of its initial value. In the context of radioactive decay, which follows similar kinetics, the half-life is the duration for half of the radioactive nuclei to undergo decay. But the concept extends to chemical reactions as well.

For a first-order reaction, the half-life is a constant value, meaning it doesn't depend on the initial concentration of the reactant. That's incredibly handy because it tells you that, no matter how much \( \mathrm{N}_{2}\mathrm{O}_{5} \) we start with, it will always take the same amount of time to reduce to half of its initial amount at a specific temperature. As demonstrated in the solution, we used the formula \( t_{1/2} = \frac{\ln{2}}{k} \) to calculate the half-life of \( \mathrm{N}_{2}\mathrm{O}_{5}\), which turned out to be 1.69 minutes at \(70^\circ \mathrm{C}\). This concept is crucial for processes that depend on precise timing of reactant concentration, like pharmaceuticals in the body or industrial chemical synthesis.